total resistance of sector of cylinder

I have sector of cylinder with inner radius $r_1$ and outer radius of $r_2$

Now, the current goes through inner radius surface to outer radius surface, it has resistivity $\rho$, and has thickness of cylinder $h$ From the top view, it looks like below.

Its central angle is $\frac{2\pi}{n}$.

Now I want to calculate the total resistance of this sector(is this how we call this figure?), what I did was

$$dR=\rho\frac{dr}{A}=\rho\frac{dr}{h2\pi r/n}$$ $$R_{tot}=\frac{n\rho}{2\pi h}\int^{r_2}_{r_1}\frac{1}{r}dr =\frac{n\rho}{2\pi h}\ln\left(\frac{r_2}{r_1}\right)$$

Is this correct?

Moreover, it was sheet resistance $R_s$. Can I replace this with resistivity $\rho=h\times R_s$?

• Appears correct. – Gert Sep 27 '16 at 23:15

Your way to calculate the resistance and your answer are correct. In this formula you can also express resistivity by sheet resistance $\rho = h \cdot R_\mathrm{s}$.